Question Statement

Evaluate the integral:

$\int \frac{1}{x^3+2x^2+x}dx$

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Background and Explanation

This problem involves integrating a rational function where the denominator is a cubic polynomial. The key technique is partial fraction decomposition, which requires factoring the denominator first. You will need to express the rational function as a sum of simpler fractions with unknown constants, then solve for those constants using substitution and coefficient comparison.

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Solution

First, factor the denominator to identify the structure of the partial fractions:

$\begin{array}{rl}\int \frac{1}{x^3+2x^2+x}dx& =\int \frac{1}{x\left(x^2+2x+1\right)}dx\\ & =\int \frac{1}{x(x+1{)}^2}dx\end{array}$

Since the denominator contains a distinct linear factor $x$ and a repeated linear factor $(x+1{)}^2$, we set up the partial fraction decomposition as:

\begin{equation*} \frac{1}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}} \end{equation*}

To find the constants $A$, $B$, and $C$, multiply both sides of equation (1) by $x(x+1{)}^2$ to clear the denominators:

\begin{equation*} 1=A(x+1)^{2}+B(x)(x+1)+C(x) \end{equation*}

Finding $A$: Substitute $x=0$ into equation (2):

$\begin{array}{rl}1& =A(0+1{)}^2+B(0)(0+1)+C(0)\\ 1& =A(1)+0+0\\ 1& =A\end{array}$

Finding $C$: Substitute $x=-1$ into equation (2):

$\begin{array}{rl}1& =A(-1+1{)}^2+B(-1)(-1+1)+C(-1)\\ 1& =A(0)+B(-1)(0)+C(-1)\\ 1& =-C\\ -1& =C\end{array}$

Finding $B$: Expand equation (2) and compare coefficients. First, expand the right side:

$\begin{array}{rl}1& =A\left(x^2+2x+1\right)+B\left(x^2+x\right)+C(x)\\ 1& =A{x}^2+2Ax+A+B{x}^2+Bx+Cx\\ 1& =(A+B)x^2+(2A+B+C)x+A\end{array}$

Comparing the coefficients of $x^2$ on both sides:

$\begin{array}{rl}0& =A+B\\ 0& =1+B(\text{since }A=1)\\ -1& =B\end{array}$

Now substitute $A=1$, $B=-1$, and $C=-1$ back into equation (1):

$\frac{1}{x(x+1{)}^2}=\frac{1}{x}+\frac{-1}{x+1}+\frac{-1}{(x+1{)}^2}=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1{)}^2}$

Integrate both sides with respect to $x$:

$\begin{array}{rl}\int \frac{1}{x(x+1{)}^2}dx& =\int \frac{1}{x}dx-\int \frac{1}{x+1}dx-\int \frac{1}{(x+1{)}^2}dx\\ & =\ln |x|-\ln |x+1|-\int (x+1{)}^{-2}dx\\ & =\ln |x|-\ln |x+1|-\frac{(x+1{)}^{-2+1}}{-2+1}+C\\ & =\ln |x|-\ln |x+1|-\frac{(x+1{)}^{-1}}{-1}+C\\ & =\ln |x|-\ln |x+1|+\frac{1}{x+1}+C\end{array}$

(Note: The absolute value bars ensure the logarithms are defined for all valid $x$ in the domain, and $C$ represents the constant of integration.)

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Key Formulas or Methods Used

• Factoring: $x^3+2x^2+x=x(x+1{)}^2$
• Partial Fraction Decomposition (for repeated linear factors): $\frac{1}{x(x+1{)}^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$
• Substitution Method: Plugging in strategic values of $x$ (roots of factors) to solve for constants
• Coefficient Comparison: Equating coefficients of like powers of $x$ after expanding
• Integration Rules:
  • $\int \frac{1}{u}du=\ln |u|$
  • $\int u^{n}du=\frac{u^{n+1}}{n+1}$ for $n\ne -1$ (power rule)

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Summary of Steps

1. Factor the denominator completely: $x^3+2x^2+x=x(x+1{)}^2$
2. Set up the partial fraction decomposition with three terms (one for $x$, two for the repeated factor $(x+1)$)
3. Clear denominators by multiplying both sides by $x(x+1{)}^2$
4. Solve for constants:
   • Substitute $x=0$ to find $A=1$
   • Substitute $x=-1$ to find $C=-1$
   • Compare $x^2$ coefficients to find $B=-1$
5. Rewrite the integrand as: $\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1{)}^2}$
6. Integrate each term separately using basic integration rules to obtain the final answer: $\ln |x|-\ln |x+1|+\frac{1}{x+1}+C$